Difference between revisions of "1998 AIME Problems/Problem 14"
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<cmath>(m' - 2) + (n' - 2) \leq (m' - 2)(n' - 2) + 1 = 9 + d.</cmath> | <cmath>(m' - 2) + (n' - 2) \leq (m' - 2)(n' - 2) + 1 = 9 + d.</cmath> | ||
So | So | ||
− | <cmath>d(8 + 4(m + n)) < 8 + 4(9 + d)</cmath> | + | <cmath>d(8 + 4(m + n)) < 8 + 4(9 + d)</cmath> //shouldn't this be 24 + 4(9+d)? then the entire proof is fallacious (it results in 2(1 + 3 + 3) < 15 which is true) |
<cmath>d(4 + 4(m + n)) < 44</cmath> | <cmath>d(4 + 4(m + n)) < 44</cmath> | ||
<cmath>d(1 + m + n) < 11</cmath> | <cmath>d(1 + m + n) < 11</cmath> |
Revision as of 18:04, 7 July 2014
Problem
An rectangular box has half the volume of an rectangular box, where and are integers, and What is the largest possible value of ?
Solution
Let’s solve for :
For the denominator, we will use a factoring trick (colloquially known as SFFT), which states that .
Clearly, we want to minimize the denominator, so . The possible pairs of factors of are . These give and respectively. Substituting into the numerator, we see that the first pair gives , while the second pair gives . We can quickly test for the denominator assuming other values to convince ourselves that is the best possible value for the denominator. Hence, the solution is .
Proof that setting the denominator to is optimal: Suppose , and suppose for the sake of contradiction that there exist such that for some and such that
This implies that
and
Substituting gives
which we rewrite as
Next, note that for to be positive, we must have and be positive, so
So
//shouldn't this be 24 + 4(9+d)? then the entire proof is fallacious (it results in 2(1 + 3 + 3) < 15 which is true)
Next, we must have that and are positive, so and . Also, by how we defined . So
a contradiction. We already showed above that there are some values of and such that that work, so this proves that one of these pairs of values of and must yield the maximal value of .
See also
1998 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.